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3 years ago

Linear inequalities What is the solution to: 4 + 8x < -20

Mathematics

2 answers:

kvasek [131]3 years ago

6 0

Answer:

x<-3

Step-by-step explanation:

4+8x<-20

4+8x-4<-20-4

8x<-24

8x/8 < -24/8

x<-3

Viktor [21]3 years ago

6 0

Hey there!

4 + 8x < -20

8x + 4 < -20

SUBTRACT 4 to BOTH SIDES

8x + 4 - 4 < -20 - 4

CANCEL out: 4 - 4 because it give you 0

KEEP: -20 - 4 because it help what number is being compared to the x-value

NEW EQUATION: 8x < -20 - 4

SIMPLIFY IT!

8x < -24

DIVIDE 8 to BOTH SIDES

8x/8 < -24/8

CANCEL out: 8/8 because it give you 1

KEEP: -24/8 because it help what number is being compared to the x-value

NEW EQUATION: x < -24/8

SIMPLIFY IT!

x < -3

IT IS AN OPEN CIRCLE SHADED to the LEFT

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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Answer:

For the sequence is $-\frac{2}{3}$ ,-4 ,-24 ,-144 ,...

Hence the formula $f(x)=-\frac{2}{3}(6)^{x-1}$ for x=1,2,3,... represents the given geometric sequence

Step-by-step explanation:

Given sequence is $-\frac{2}{3}$ ,-4 ,-24 ,-144 ,...

To find the formula to describe the given sequence :

Let $a_1=\frac{-2}{3}$ , $a_2=-4$ , $a_3=-24$ ,...

First find the common ratio

$r=\frac{a_2}{a_1}$ here $a_1=\frac{-2}{3}$ and, $a_2=-4$

$=\frac{-4}{\frac{-2}{3}}$

$=\frac{4\times 3}{2}$

$=\frac{12}{2}$

$r=6$

$r=\frac{a_3}{a_2}$ here $a_2=-4$ and $a_3=-24$

$=\frac{-24}{-4}$

$=6$

$r=6$

Therefore the common ratio is 6

Therefore the given sequence is geometric sequence

The nth term of the geometric sequence is

$a_n=a_1r^{n-1}$

The formula which describes the given geometric sequence is

$f(x)=a_1r^{x-1}$ for x=1,2,3,...

$=\frac{-2}{3}6^{x-1}$ for x=1,2,3,...

Now verify that $f(x)=a_1r^{x-1}$ for x=1,2,3,... represents the given geometric sequence or not

put x=1 and the value of $a_1$ in $f(x)=a_1r^{x-1}$ for x=1,2,3,...

we get $f(1)=-\frac{2}{3}(6)^{1-1}$

$=-\frac{2}{3}(6)^0$

$=-\frac{2}{3}$

Therefore $f(1)=-\frac{2}{3}$

put x=2 we get $f(2)=-\frac{2}{3}(6)^{2-1}$

$=-\frac{2}{3}(6)^1$

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Therefore $f(2)=-4$

put x=3 we get $f(3)=-\frac{2}{3}(6)^{3-1}$

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$=-\frac{2(36)}{3}$

Therefore $f(3)=-24$

Therefore the sequence is f(1),f(2),f(3),...

Therefore the sequence is $-\frac{2}{3}$ ,-4 ,-24 ,-144 ,...

Hence the formula $f(x)=a_1r^{x-1}$ for x=1,2,3,... represents the given geometric sequence is verified

Therefore the formula $f(x)=-\frac{2}{3}(6)^{x-1}$ for x=1,2,3,... represents the given geometric sequence

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